Mapping an arc to a line segment while keeping the infinity The original problem is to map the complement of the arc $|z|=1, y\ge 0$ on the outside of the unit circle, so that the points at $\infty$ correspond to each other. (with a conformal map)
My idea is to map the arc to the line segment $|x|\le 1,y=0$, then use the function $w=z+\sqrt{z^2-1}$. Since the latter function keeps $\infty$, the former function also has to keep $\infty$. But this seems impossible: to map an arc to a line segment using a Mobius function, a point on the circle must be sent to $\infty$, so the original $\infty$ is sent to a finite point. A logarithmic function does the job, but it is not defined on $0$.
Am I in the wrong direction?
 A: The function that you are looking for can be obtained in a few steps.
For pedagogical purposes, I will first give a wrong solution below.
First, take the cross ratio $f_1(z) = (z,1,-1, -i)$. This maps the unit circle to the real axis. In particular, it maps the arc in question to the line segment $[0, 1]$. Also, it maps $\infty$ to $\frac{1+i}{2}$.
Then take $f_2(z) = z^4$. $f_2 \circ f_1$ maps the arc to $[0, 1]$ and $\infty$ to $-\frac{1}{4}$.
The rest is a translation moving $-\frac{1}{4}$ to 0, an inversion so that 0 becomes $\infty$, and another translation/scaling to take the line segment to [-1,1].
The problem with this solution is that with $z^4$, four points (so not only $\frac{1+i}{2}$) will be mapped to $-\frac{1}{4}$. This means there are three other points that are mapped to $\infty$ in the final result. So the correspondence to $\infty$ is not one-to-one. In each step, you want to have a function that is bijective.
Below is a right solution.
First, take a cross ratio (possibly with translation and scaling) to map the arc to [-1, 1], and $\infty$ to $a$ (which is certainly not on the real axis). Let us denote this function as $f_1(z)$.
Second, use the function $f_2(z) = z-\sqrt{z^2-1}$, where sign of the square root is uniquely determined by the condition $|f(z)| < 1$. Let $f(a) = b$.
Third, take $f_3(z) = \frac{b-z}{1-\bar{b}z}$. This is an automorphism of the unit disc that swaps 0 and $b$.
Finally, take $f_4(z) = \frac{1}{z}$. This function swaps the interior of the unit disc with the exterior, and maps 0 to $\infty$.
$f_4 \circ f_3 \circ f_2 \circ f_1$ is the desired mapping.
